When an airplane is 35,000 feet from an air traffic control tower, the angle of elevation between the tower and the airplane is. What is the approximate altitude, a, of the plane at this point?
A. 45,689 feet
B. 26,812 feet
C. 29,369 feet
D. 41,711 feet

26,812 ft

Step-by-step explanation:

The drawing given is very helpful in this case.  When solving problems like this, it's important to realize what trigonometric ratio we're going to use.  From the given angle (50°), we are given the hypotenuse (35,000 ft) and we're trying to solve for the opposite side ().  

So since we're trying to find the opposite side side and we have the hypotenuse, we should try to find a trigonmetric ratio between the opposite side and the hypotenuse.  Using SOH-CAH-TOA, hopefully you can see that we should pick SOH (i.e. ).  Therefore, we can set up our equation given the angle .  

Since we're solving for , we can just rearrange to get

Therefore, the plane's altitude is 26,812 ft.

m∠cdf = 43° , m∠gdf = 114°, m∠gde = 23°, m∠edh = 23° , m∠cdh = 157°

step-by-step explanation:

since, it is given that de bisects the angle gdh ⇒ de divides the angle gdh into two equal angles.

⇒ m∠gde = m∠edh

⇒ 8·x   - 1 = 6·x + 5

⇒ x = 3

so, m∠gde = 8·x - 1 = 23°

and m∠edh = 23°

now, it is given ∠cde is an straight angle so, m∠cde = 180°

⇒ m∠cdf + m∠fdg + m∠gde = 180°

⇒ 43° + m∠fdg + 23° = 180°

⇒ m∠fdg = 114°

again, m∠cde = m∠cdh + m∠edh = 180°

m∠cdh + 23° = 180°

⇒ m∠cdh = 157°

translate k to n and reflect across the line containing jk.